[CODE](18:10)>C= vector(100,n,(2*n)!/((n+1)!*n!))
%10 = [1, 2, 5, 14, 42, 132, 429, 1430, 4862,
(18:10)>Catalan(n) = return(C[n+1])
%11 = (n)->return(C[n+1])
(18:12)>Catalan(0)
%12 = 1
(18:12)>Catalan(2)
%13 = 5[/CODE]

does this cover

[url]http://rosettacode.org/wiki/Catalan_numbers[/url] ?

I forgot [url]http://www.mersenneforum.org/showthread.php?p=248234#post248234[/url] lead to:

[url]http://rosettacode.org/wiki/Luhn_test_of_credit_card_numbers[/url]

done it should be put up.

sorry for getting off topic again I see a few others that shouldn't be too difficult in the list but anyways.

[QUOTE=science_man_88;252712][CODE](18:10)>C= vector(100,n,(2*n)!/((n+1)!*n!))
%10 = [1, 2, 5, 14, 42, 132, 429, 1430, 4862,
(18:10)>Catalan(n) = return(C[n+1])
%11 = (n)->return(C[n+1])
(18:12)>Catalan(0)
%12 = 1
(18:12)>Catalan(2)
%13 = 5[/CODE]

does this cover

[url]http://rosettacode.org/wiki/Catalan_numbers[/url] ?[/QUOTE]

Sure, that'll do -- as long as you don't want Catalan(100) or higher. (Edit: I see that the link suggests memoization, in which case you did it right. As it happens it's probably not needed here.) I would just do
[code]Catalan(n)=(2*n)!/(n+1)!/n![/code]
or, probably faster,
[code]Catalan(n)=binomial(2*n,n+1)/n[/code]

[QUOTE=science_man_88;252720]I forgot [url]http://www.mersenneforum.org/showthread.php?p=248234#post248234[/url] lead to:

[url]http://rosettacode.org/wiki/Luhn_test_of_credit_card_numbers[/url]

done it should be put up.[/QUOTE]

Great, put it up!

[QUOTE=science_man_88;252736]sorry for getting off topic again I see a few others that shouldn't be too difficult in the list but anyways.[/QUOTE]

You can do them if you like, or post it here if you want feedback.

[QUOTE=CRGreathouse;252751]Great, put it up!



You can do them if you like, or post it here if you want feedback.[/QUOTE]

well I kinda see the math of the sierpinski triangle and carpet

going up a step from the triangle shown means multiplying the width by 3 and printing 8/9 of the previous carpet and leaving the center clear. since they show examples with them hitting the smallest possible it's simply working backwards I think.

[CODE](20:28)>sdsub(x,y) = C=y;for(a=1,#C,C[a]=x[C[a]+1]);C=vecsort(C);for(b=1,#y,x[y[b]+1]=C[b]);x;
(20:48)>sdsub([7, 6, 5, 4, 3, 2, 1, 0],[6, 1, 7])
%94 = [7, 1, 5, 4, 3, 2, 0, 6][/CODE] this is trying :
[url]http://rosettacode.org/wiki/Sort_disjoint_sublist[/url]

it seems to reverse 2 thing as it places them. oh never mind I see why trying out the simpler solution they suggest, it's because the call to a unsorted y.

I think this is what PARI's indirect sorting is for. Take the elements you want to sort into a new vector, call it u, and sort with vecsort(u,,1). Rather than sorting u, this will return a vector that tells you where to put the elements. Then you can just put each element where the resulting vector tells you to put it.

[QUOTE=CRGreathouse;252861]I think this is what PARI's indirect sorting is for. Take the elements you want to sort into a new vector, call it u, and sort with vecsort(u,,1). Rather than sorting u, this will return a vector that tells you where to put the elements. Then you can just put each element where the resulting vector tells you to put it.[/QUOTE]

[CODE](21:28)>vecsort([1,2,3],,1)
%134 = Vecsmall([1, 2, 3])
(21:30)>vecsort([1,2,3],,2)
%135 = [1, 2, 3]
(21:30)>vecsort([1,2,3],,3)
%136 = Vecsmall([1, 2, 3])
(21:33)>vecsort([1,2,3],,4)
%137 = [3, 2, 1]
(21:33)>vecsort([1,2,3],,5)
%138 = Vecsmall([3, 2, 1])
(21:33)>vecsort([1,2,3],,6)
%139 = [3, 2, 1]
(21:33)>vecsort([1,2,3],,7)
%140 = Vecsmall([3, 2, 1])
(21:33)>vecsort([1,2,3],,8)
%141 = [1, 2, 3]
(21:33)>vecsort([1,2,3],,9)
%142 = Vecsmall([1, 2, 3])
(21:34)>vecsort([1,2,3],,10)
%143 = [1, 2, 3][/CODE]

[CODE](21:25)>sdsub1(x,y) = D=y;C=y;for(a=1,#C,C[a]=x[C[a]]);C=vecsort(C);D=vecsort(D);for(b=1,#y,x[D[b]]=C[b]);x;
(21:28)>sdsub1([7, 6, 5, 4, 3, 2, 1, 0],[7,2,8])[/CODE]

okay this works for the 1 based what if I want it to act like it's 0 based.

[QUOTE=science_man_88;252865]%134 = Vecsmall([1, 2, 3])
(21:30)>vecsort([1,2,3],,2)
%135 = [1, 2, 3]
(21:30)>vecsort([1,2,3],,3)
%136 = Vecsmall([1, 2, 3])
(21:33)>vecsort([1,2,3],,4)
%137 = [3, 2, 1]
(21:33)>vecsort([1,2,3],,5)
%138 = Vecsmall([3, 2, 1])
(21:33)>vecsort([1,2,3],,6)
%139 = [3, 2, 1]
(21:33)>vecsort([1,2,3],,7)
%140 = Vecsmall([3, 2, 1])
(21:33)>vecsort([1,2,3],,8)
%141 = [1, 2, 3]
(21:33)>vecsort([1,2,3],,9)
%142 = Vecsmall([1, 2, 3])
(21:34)>vecsort([1,2,3],,10)
%143 = [1, 2, 3][/CODE][/QUOTE]

Not ,,2. Not ,,3. The flag you want to set is 1 and none of the others. (You can see what they do by reading the manual, or get an overview with ?vecsort, but they're not applicable here.)